Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 7


0 and 2.

Work Step by Step

$ f'(x)= -x^{-2}+2x^{-3}=\frac{2}{x^{3}}-\frac{1}{x^{2}}$ Since the denominator can't be 0, $ f'(0)$ doesn't exist. That is, 0 is a critical point of $ f $. Another critical point is the solution of $ f'(x)=0$ That is, $\frac{2}{x^{3}}-\frac{1}{x^{2}}=0$ $\implies \frac{2}{x^{3}}=\frac{1}{x^{2}}$ Or: $2=\frac{x^{3}}{x^{2}}=x $ Thus, the critical points are x=0, 2.
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